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kramer
2 years ago
6

A model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the ca

r and the speed of the car
​
Physics
1 answer:
valkas [14]2 years ago
4 0

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

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How are interference patterns made?
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To set up a stable and clear interference pattern, two conditions must be met: The sources of the waves must be coherent, which means they emit identical waves with a constant phase difference.
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3 years ago
A submarine has a "crush depth" (that is, the depth at which
sergey [27]

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm

So, the approximate pressure is equal to 40.69 atm.

3 0
2 years ago
Two similar gold spheres are separated by a distance of
solong [7]

Answer:

The answer to your question is the letter D. 2.5 N

Explanation:

The electrostatic force is the same in both directions,

If the electrostatic force on B due to A is 2.5 N, the magnitude of the electrostatic force on A due to B must be 2.5N.

Maybe the direction is different but the magnitude is the same.

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3 years ago
Devon places sliced potatoes in a package of aluminum foil. He puts a metal grate over a campfire and sets the package on the gr
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when he set the potatoes in the aluminum foil over the campfire because the heat was directly touching them and heating them

7 0
3 years ago
Read 2 more answers
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
2 years ago
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